d electrolysis of CuSO4(aq) using platinum electride yields Cu2+ deposit at d cathode and O2 2- at d anode.
As the Cu2+ and OH- are gradually discharged, d solution becomes acidic wit H2SO4 due to the H+ and SO4 2- wich remain in the solution,This change is seen by the gradual fading away of the blue color of CuSO4.If we continue to pass a current trough d solution,subsequent electrolysis of this dilute acid will occur,yielding H at d cathode and O2 at the anode

toeseen:
by the hydrolysis of cuso4..cu+,h+,so42-,oh-...are the ions, the anode is the positive electrode which is atrracted to the negative ions i.e the anions, oh- is thus deposited because it is platinum electrode and higher in the ecs,and since oh- is a basic ion, alkali ion..ot turns red litmus blue...and oxidation occurs at the anode to liberate oxygen gas!

at the anode, the hydroxyl ion is discharged and we have.... 4OH-->2H2O+O2. solution is acidic by the presence of H+ and SO4^2- ions combined into H2SO4. therefore, the litmus turns red and oxygen is evolved..

The correct answer is B. At the anode OH- is given off which combines with itself to give of oxygen Hence the presence of OH- ions makes red litmus applied to the anode to turn blue due to its alkalinity.